Recall that \(f = \sum _{n \ge 1} a_n(E)q^n\)  admits a mod \(p\)  companion form if
there exists a mod \(p\)  eigenform
\(g = \sum _{n \ge 1} b_n q^n\)  of weight \(p-1\) 
such that
\(n^2 \cdot b_n \equiv n \cdot a_n(E) \pmod {p}\) 
for all \(n \ge 1\) .
By [1]}, the above diagonalizability is also equivalent to the existence of a mod \(p\)  companion form of \(f\) .
